Lipschitz constant tending to zero implies function is constant

Question

Let $f$ be a Lipschitz function defined on a ball $B$ in $\mathbb{R}^n$, and assume that the following property on the Lipschitz constant (restricted to smaller balls) holds:

$$ \lim_{r\to 0}\sup_{|x-y|\leq r} \frac{|f(x)-f(y)|}{|x-y|}=0. $$

Can we assert that $f$ is constant? Intuitively, this should be the case, since this limit should tend to the supremum of the derivative. However, I was not able to fill in the details.

Answer 1

My thanks go to @Bruno B for pointing out an error in my argument.

Let's fix $x$ and consider the limit as $y$ approaches $x$. Since $y \to x$ as $|x-y| \to 0$, we can parametrize $y$ by $y=x+r\theta$, where $r$ is the distance $|x-y|$ and $\theta$ is some angle in $\mathbb{R}^n$ (see n-DIMENSIONAL SPHERE)

For any $r$ we have $$\sup_{θ} \frac{|f(x)-f(x+r\theta)|}{|x-(x+r\theta)|}=\sup_{y:\;|x-y|= r} \frac{|f(x)-f(y)|}{|x-y|}\leq \sup_{y:\;|x-y|\leq r} \frac{|f(x)-f(y)|}{|x-y|}$$ Therefore, \begin{align} 0&\leq\lim_{y \to x} \frac{|f(x)-f(y)|}{|x-y|}\\ &=\lim_{|y-x| \to 0} \frac{|f(x)-f(y)|}{|x-y|}\\ &=\lim_{r \to 0}\frac{|f(x)-f(x+r\theta)|}{|x-(x+r\theta)|}\\ &\leq \lim_{r\to 0}\left(\sup_{θ} \frac{|f(x)-f(x+r\theta)|}{|x-(x+r\theta)|}\right)\\ &\leq \lim_{r\to 0}\left(\sup_{y:\;|x-y|\leq r} \frac{|f(x)-f(y)|}{|x-y|}\right)\\ &= 0. \end{align} By the definition of the derivative, this limit is equivalent to $|f'(x)|$, thus we have: $$ \forall x\in B,\quad |f'(x)| = 0. $$ We can conclude that $f$ is constant on the entire ball $B$.

Answer 2

Let $x \in B$ be fixed. Then, for $r > 0$ small enough that $x + h$ belongs in $B$ when $|h| \leq r$, we have: $$0 \leq \frac{|f(x+h) - f(x)|}{|h|} \leq \sup_{h',\, |h'| \leq |h|} \frac{|f(x+h') - f(x)|}{|h'|}$$ Taking the $\limsup_{|h| \to 0}$ on this inequality, we obtain: $$\begin{split} 0 \leq \limsup_{|h| \to 0} \frac{|f(x+h) - f(x)|}{|h|} &\leq \limsup_{|h| \to 0} \sup_{h',\, |h'| \leq |h|} \frac{|f(x+h') - f(x)|}{|h'|}\\ &\leq \limsup_{r \to 0}\sup_{h',\, |h'| \leq r} \frac{|f(x+h') - f(x)|}{|h'|} = 0\end{split}$$ where the last inequality (which is an equality) holds thanks to the fact that the expression only depends on $|h|$ and not $h$.
Hence $$\limsup_{|h| \to 0} \underbrace{\frac{|f(x+h) - f(x)|}{|h|}}_{\geq 0} = 0$$ and thus $\liminf = \limsup = 0$, meaning that the limit exists and is equal to: $$\lim_{|h| \to 0} \frac{|f(x+h) - f(x)|}{|h|}= 0$$ Finally, this means that the differential of $f$ at $x$ is the zero functional, therefore, since $B$ is connected and this holds for all $x \in B$, $f$ is constant.

Answer 3

The other answers are fine but I would like to point out that this result is also true in much greater generality, and with a non-uniform assumption on the limit, provided we simply satisfy a rectifiability condition.

Namely, let $X$, $Z$ be metric spaces, and let $f\colon X\to Z$ satisfy the infinitesimal Lipschitz condition that for each $x\in X$, we have $$\lim_{r\to 0}\sup_{y\in X \text{, } d(x,y)\leq r} \frac{d(f(x),f(y))}{d(x,y)}=0\text{.} $$

Then $f$ is constant on every rectifiable curve in $X$. In particular, if $X$ is rectifiably connected, then $f$ is constant.

Proof

The idea here is we combine rectifiability and the point-separating nature of Lipschitz functions to reduce to the real case.

That is, on every rectifiable path $\gamma\colon[0,l(\gamma)]\to X$, parametrized by arc-length, and every $1$-Lipschitz function $L$ on $Z$, we have at each $s_0\in[0,l(\gamma)]$ that

\begin{align*} |(L\circ f\circ\gamma)(s)-(L\circ f\circ\gamma)(s_0)| &\leq d((f\circ\gamma)(s),(f\circ\gamma)(s_0))\\ &\leq \omega(d(\gamma(s),\gamma(s_0))) \cdot d(\gamma(s),\gamma(s_0))\\ &\leq \omega(d(\gamma(s),\gamma(s_0))) \cdot |s-s_0|\text{,} \end{align*} where $\omega(\delta)\to 0$ as $\delta\to 0$, whereby $(L\circ f\circ \gamma)'(s_0)=0$. Thus $L\circ f\circ \gamma$ is constant on $[0,l(\gamma)]$, so that if $p,q \in \gamma([0,l(\gamma)])$, $L(f(p))=L(f(q))$. Applying this to the $1$-Lipschitz function $L(z)=d(z,f(p))$, we get that $$0=d(f(p),f(p))=d(f(q),f(p))\text{,}$$ and so $f(p)=f(q)$.

Remark

Without a condition like rectifiable connectedness, there are simple counterexamples to this result, e.g., let $X=[0,1]$ with the "snowflake" metric $d(x,y)=|x-y|^{\frac{1}{2}}$, and let $Z=\mathbb R$ with the usual metric, then let $f$ be the identity.